# Leetcode Algorithms

448. Find All Numbers Disappeared in an Array

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

`Input:[4,3,2,7,8,2,3,1]Output:[5,6]`

Logic:

Firstly, the prerequisite is 1 ≤ a[i] ≤n; secondly, we compare nums[i] and nums[nums[i]-1]. If they are not the same, swap them. And then we validate the res vector by checking nums[i] and i+1. If they are not the same, then we store i+1 in the res vector

Solution:

`class Solution{    public:        vector<int> findDisappearedNumbers(vector<int>&nums){            vector<int> res;            for (int i = 0; i < nums.size(); ++i){                if (nums[i] != nums[nums[i] - 1]){                    swap(nums[i], nums[nums[i] - 1]);                    -- i;                }            }            for (int i = 0; i < nums.size(); ++i){                if (nums[i] != i + 1){                    res.push_back(i + 1);                }            }            return res;    }};`

Reference

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